Theorem 12 (Gromov): Let be torsion-free -hyperbolic group. If such that , then for all sufficiently large , .
Remark: The torsion-free hypothesis is not necessary, but it allows us to avoid some technicalities. For instance, it is a non-obvious fact that an infinite hyperbolic group contains a copy of .
For the rest of this lecture will be a torsion-free -hyperbolic group, where are primitive (i.e. not proper powers).
Recall that for torsion-free -hyperbolic, primitive implies that .
If and do not commute we can show there is some point on arbitrarily far from .
Hence we have the following lemma.
If and do not commute there is some point on arbitrarily far from .
Proof: Suppose not. That means such that such that . So is in . But the Cayley graph is locally finite so has finitely many elements. By the Pigeonhole Principle such that for some . Then . But then . .
For a moment view and as the horizontal and vertical geodesics in . For two points on and on , we can argue that the geodesic between them curves toward the origin.
And so we have Lemma 14.
Lemma 14: There exists such that , .
Recall that by is a quasi-isometric embedding. So by Theorem 6, and
By Lemma 13 choose such that
. Choose such that . Now, must be -close to so for some point on the geodesic between and , . Then .
For a subgroup , one can choose a closest point projection which is -equivariant. (Write . Choose where and are close and declare to be -equivariant.) is typically not a group homomorphism.
We’re interested in and .
In , there is some such that either or .
Lemma 15: such that , or .
Let . WLOG, is -close to and since is the closest point to (in particular compared to ). So . .
Now we can prove the theorem.
Proof of Theorem 12:
The idea is to use the Ping-Pong Lemma on the Cayley graph.
Let and let , where is provided by Lemma 15. For all we have and likewise for all we have . In particular, .
Let . By -equivariance,
for any . In particular,
by the triangle inequality. Similarly,
for all and all . Because and are quasi-isometrically embedded, it follows that and for .
Therefore, by the Ping-Pong Lemma .